Trigonometry Equations

Learn to solve sin, cos, and tan equations step by step

CAPS Grade 10 Mathematics

Trigonometric equations involve finding angles that make the equation true. Use the unit circle and reference angles to find all solutions within a given range.

What You'll Learn

Solve sin θ = k, cos θ = k, tan θ = k
Use the unit circle to find solutions
Find reference angles for any angle
Find all solutions from 0° to 360°
Write general solutions with k·360° or k·180°
Solve equations that need algebra first

Quiz 1: ASTC Rule

In which quadrant is tan positive

A) Quadrant I only
B) Quadrant II only
C) Quadrant I and III
D) Quadrant II and IV

1. Unit Circle & Reference Angles

U

Unit Circle

  • Circle with radius = 1
  • Center at (0,0)
  • sin θ = y-coordinate
  • cos θ = x-coordinate
  • tan θ = y/x = sin θ/cos θ
R

Reference Angle

  • Acute angle (0° to 90°)
  • Angle between terminal arm and x-axis
  • Used to find trig values in all quadrants
A

ASTC Memory Aid

  • All positive in QI
  • Sin positive in QII
  • Tan positive in QIII
  • Cos positive in QIV
QuadrantRangeSinCosTanReference Angle
I0°-90°+++θ
II90°-180°+--180° - θ
III180°-270°--+θ - 180°
IV270°-360°-+-360° - θ

Quiz 2: Solving sin θ = 0.5

What are the solutions for sin θ = 0.5 between 0° and 360°

A) 30°, 150°
B) 30°, 330°
C) 60°, 120°
D) 45°, 135°

2. Solving sin θ = k

Step-by-Step: sin θ = 0.5

Example

Solve sin θ = 0.5 for 0° ≤ θ ≤ 360°

Solution Steps
  1. Reference angle: sin⁻¹(0.5) = 30°
  2. Sin is positive in Quadrants I and II
  3. QI: θ = 30°
  4. QII: θ = 180° - 30° = 150°
  5. Both are within 0°-360°
Answer: θ = 30°, 150°

Example: sin θ = -0.5

Example

Solve sin θ = -0.5 for 0° ≤ θ ≤ 360°

Solution Steps
  1. Reference angle: sin⁻¹(0.5) = 30° (use absolute value)
  2. Sin is negative in Quadrants III and IV
  3. QIII: θ = 180° + 30° = 210°
  4. QIV: θ = 360° - 30° = 330°
Answer: θ = 210°, 330°

Quiz 3: Solving cos θ = 0.5

What are the solutions for cos θ = 0.5 between 0° and 360°

A) 30°, 150°
B) 30°, 330°
C) 60°, 120°
D) 60°, 300°

3. Solving cos θ = k

Step-by-Step: cos θ = 0.5

Example

Solve cos θ = 0.5 for 0° ≤ θ ≤ 360°

Solution Steps
  1. Reference angle: cos⁻¹(0.5) = 60°
  2. Cos is positive in Quadrants I and IV
  3. QI: θ = 60°
  4. QIV: θ = 360° - 60° = 300°
Answer: θ = 60°, 300°

Example: cos θ = -0.5

Example

Solve cos θ = -0.5 for 0° ≤ θ ≤ 360°

Solution Steps
  1. Reference angle: cos⁻¹(0.5) = 60°
  2. Cos is negative in Quadrants II and III
  3. QII: θ = 180° - 60° = 120°
  4. QIII: θ = 180° + 60° = 240°
Answer: θ = 120°, 240°

Quiz 4: Solving tan θ = 1

What are the solutions for tan θ = 1 between 0° and 360°

A) 30°, 210°
B) 45°, 135°
C) 45°, 225°
D) 60°, 240°

4. Solving tan θ = k

Step-by-Step: tan θ = 1

Example

Solve tan θ = 1 for 0° ≤ θ ≤ 360°

Solution Steps
  1. Reference angle: tan⁻¹(1) = 45°
  2. Tan is positive in Quadrants I and III
  3. QI: θ = 45°
  4. QIII: θ = 180° + 45° = 225°
Answer: θ = 45°, 225°

Example: tan θ = -1

Example

Solve tan θ = -1 for 0° ≤ θ ≤ 360°

Solution Steps
  1. Reference angle: tan⁻¹(1) = 45°
  2. Tan is negative in Quadrants II and IV
  3. QII: θ = 180° - 45° = 135°
  4. QIV: θ = 360° - 45° = 315°
Answer: θ = 135°, 315°

Quiz 5: Algebra First

Solve 2sin θ + 1 = 0 for 0° ≤ θ ≤ 360°

A) 30°, 150°
B) 210°, 330°
C) 60°, 300°
D) 120°, 240°

5. Algebra First

Step-by-Step: 2sin θ + 1 = 0

Example

Solve 2sin θ + 1 = 0 for 0° ≤ θ ≤ 360°

Solution Steps
  1. Isolate sin θ: 2sin θ = -1
  2. Divide by 2: sin θ = -0.5
  3. Reference angle: sin⁻¹(0.5) = 30°
  4. Sin negative in QIII and QIV
  5. QIII: θ = 180° + 30° = 210°
  6. QIV: θ = 360° - 30° = 330°
Answer: θ = 210°, 330°

Example: 3cos θ - 2 = 0

Example

Solve 3cos θ - 2 = 0 for 0° ≤ θ ≤ 360°

Solution Steps
  1. Isolate cos θ: 3cos θ = 2
  2. Divide by 3: cos θ = 2/3 ≈ 0.6667
  3. Reference angle: cos⁻¹(2/3) ≈ 48.19°
  4. Cos positive in QI and QIV
  5. QI: θ ≈ 48.19°
  6. QIV: θ ≈ 360° - 48.19° = 311.81°
Answer: θ ≈ 48.19°, 311.81°

6. General Solutions

General Solution Formulas
sin θ = k: θ = a + k·360° or θ = (180°-a) + k·360°
cos θ = k: θ = a + k·360° or θ = -a + k·360°
tan θ = k: θ = a + k·180°

General Solution Example

Example

Find the general solution for sin θ = 0.5

Solution
Reference angle: a = 30°
θ = 30° + k·360° or θ = 150° + k·360°, k ∈ ℤ

For k=0: 30°, 150°

For k=1: 390°, 510°

For k=-1: -330°, -210°

Quiz 6: General Solution

What is the general solution for tan θ = √3

A) θ = 30° + k·360°
B) θ = 60° + k·180°
C) θ = 60° + k·360°
D) θ = 30° + k·180°

7. Common Mistakes

Mistake:

Wrong quadrant selection - always use ASTC: All, Sin, Tan, Cos positive

Mistake:

Using signed value for reference angle - reference angle is always acute (0°-90°)

Mistake:

Forgetting to check domain - ensure solutions are within the given range

Mistake:

Wrong period for general solution - sin/cos use 360°, tan uses 180°

8. Practice Questions

Q1

Solve sin θ = √3/2 for 0° ≤ θ ≤ 360°

Answer: 60°, 120°
Q2

Solve cos θ = -√2/2 for 0° ≤ θ ≤ 360°

Answer: 135°, 225°
Q3

Solve 2tan θ - 2√3 = 0 for 0° ≤ θ ≤ 360°

Answer: 60°, 240°

9. Summary

Solution Formulas

  • sin θ = k: θ = a or 180° - a
  • cos θ = k: θ = a or 360° - a
  • tan θ = k: θ = a or 180° + a

General Solutions

  • sin: θ = a + k·360° or (180°-a) + k·360°
  • cos: θ = ±a + k·360°
  • tan: θ = a + k·180°

ASTC Rule

  • QI: All positive
  • QII: Sin positive
  • QIII: Tan positive
  • QIV: Cos positive
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